So I built this Utility Distribution module cobbled together from various circuits.... I have gate and trigger buffered mults that are Ray Wilson's design, I have a CV mult with no offsets based on Ray's design but with the glide and offset omitted, another CV mult with offsets but no glide, same source. Then I wanted to do Sum and Difference. So I have A and B jacks with attenuation, and sum and difference circuits taken more or less from textbook examples (I don't remember precisely, but the difference circuit was probably from Forrest M Mims Op Amp Notebook).

Except I have found that the sum and difference don't play nice together using the same inputs, at least as I have naively put them together.

I sum the two with a simple standard mixer circuit. But then I also take the attenuated B input and use it as negative input to another op amp, and the attenuated A input and use it as the positive input. See the attached schematic fragment.

It's not shown, but I'm using nc jacks with the normal connection grounded. Due to the way these are wired, when I'm doing A = CV and B = 0 (nothing plugged in) the B attenuator actually creates a voltage divider from A through the two summing input resistors and the parallel resistances on either side of the wiper, so that I get a nonzero B even when both sides of the pot are grounded--unless I have the pot all the way at either extreme. I get similar results when I'm trying to do A = 0 and B = CV to get an inversion.

I've googled and searched here for any actual application circuits but haven't turned anything up. I'd think maybe a strategically placed diode might work, except that I'd then be contending with the diode drop (sum a diode drop back in to compensate?). So am I right in thinking the only thing I can do to fix this is a buffer between the attenuators and the sum/difference input resistors?

Silly me for thinking "oh this is so simple I don't need to breadboard it"

Here's an edited version of the schematic, showing where the voltage divider current I'm talking about is going. Keep in mind that in the first example, the B input (tip) his held at ground as well as the bottom of the attenuation pot.

However with that in mind, I could make the summing amp inputs 1M and that *should* have the desired effect. At it's highest the resistance of the parallel sides of the pot will be 25k (a middle-split, with 50K on either side), which with 100k summing resistors leads to a voltage divider of 1/9--that's consistent with what I saw, input of 5V at A leading to an input of roughly .4V at B. If I make the summing resistors 1M then it would be 1/8001, so that might work.

I would say that it should work correctly as it is right now.
You should achieve a zero when no signal is applied.
And I don't see any voltage divider. Keep in mind that the inverting input of IC4A is virtual ground. The opamp makes the two inputs be at same level. Since +input is grounded, -input is also 0.

I would look for any short in the PCB, or any mistake in the PCB layout.
By the way, there is no need to ground the normal connection._________________electronic-sea.net

Ok, I was neglecting the virtual ground at IC4A pin 2. It's acting as I describe though, perhaps my connection on pin 3 is bad....though I'd expect it not to work at all then

When you speak of the normal connection, do you mean R8 on the + input of the difference op amp?

Hm....

Let's look at this a different way. Voltage A at J1 tip, through atten pot which is for sake of argument all the way up, then goes through 100K to IC1A pin 3 (by the way, the chip indicators are not accurate between the diagram and the actual wiring, it's likely more like IC4B). And then through another 100K to ground. So *there* is a voltage divider as well, and that would actually hold pin 2 at the same potential as pin 3, and then THAT going through 100K and B's attenuation pot ... could that be the problem? But then changing B's attenuation pot shouldn't have any effect...

I will Check IC4A pin 3 and remove R8 and see if anything changes, and then go looking for other problems.

I think this is the problem. The parallel resistance from the wiper to ground of the lower pot and R6 form a voltage divider attached to the the inverting input of IC1A, the inverting input will be at the same voltage as the non inverting input. What ever voltage is on the wiper of the pot then gets fed into the summer through R3.
So worse case, with the pot in the middle and 5 volts on the inverting input you get 5 x 50/150 = 1.666 volts going into the summer. That's why I think making R6 bigger will reduce the problem.

When you speak of the normal connection, do you mean R8 on the + input of the difference op amp?

I was referring to this that you said:

Quote:

It's not shown, but I'm using nc jacks with the normal connection grounded

About the differential amplifier, IC1A, you can analysis it grounding first R6 and after R7, then you sum the results and you get the substractor. For the positive input (with R6 grounded), the voltage divider R7-R8 halves the signal but then is doubled by R19-R6, so the gain is 1. For the negative input (with R7 grounded) it is just an inverting amplifier with gain-1._________________electronic-sea.net

It's not shown, but I'm using nc jacks with the normal connection grounded. Due to the way these are wired, when I'm doing A = CV and B = 0 (nothing plugged in) the B attenuator actually creates a voltage divider from A through the two summing input resistors and the parallel resistances on either side of the wiper, so that I get a nonzero B even when both sides of the pot are grounded--unless I have the pot all the way at either extreme

Ah! now I have understood! So you increase the resistors as slacker said or decrease the resistance of the pots to minimize the effect.

Better would be to buffer the signals before split them. That are two extra op-amps. Maybe you could do it only with one more opamp?_________________electronic-sea.net

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